Question

If $x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}}$, value of $x^2+x-1$ is

• A. $\sqrt{5}$
• B. $2$
• C. $-\sqrt{5}$
• D. $\sqrt{5}+1$

Answer 1

The correct option is D $\sqrt{5}+1$

$x=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}}=\sqrt{\frac{\sqrt{5}+1}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}}=\sqrt{\frac{(\sqrt{5}+1{)}^2}{5-1}}=\frac{\sqrt{5}+1}{2}\therefore x^2+x-1={\left(\frac{\sqrt{5}+1}{2}\right)}^2+\frac{\sqrt{5}+1}{2}-1=\frac{5+1+2\sqrt{5}}{4}+\frac{\sqrt{5}+1}{2}-1=\frac{3+\sqrt{5}}{2}+\frac{\sqrt{5}+1}{2}-1=\frac{3+\sqrt{5}+\sqrt{5}+1-2}{2}=\sqrt{5}+1$